Question

In the given figure, $P_1$ and $P_2$ are the pressure experienced at the points $1$ and $2$ on a charged conductor respectively. Then:

• A. $P_1>P_2$
• B. $P_1<P_2$
• C. $P_1=P_2$
• D. can't be interpreted

Answer 1

The correct option is A $P_1>P_2$



From the figure we can see that, radius of the surface at point $P_1$ is greater then the radius of the surface at point $P_2$ i.e. $R_1>R_2$.

Therefore, electric charge density of the surface at point $P_1$ is greater then that of $P_2$ i.e. ${\sigma }_1>{\sigma }_2$.

We know that the pressure on a charged surface is given by :
$P=\frac{{\sigma }^2}{2{ϵ}_o}$

$\Rightarrow P\propto {\sigma }^2$

Here, ${\sigma }_1>{\sigma }_2$

$\therefore P_1>P_2$